MHB -2.2.4 find general solution xy'+2y=e^x

[karush]

$\textsf{find the general solution}$
$$xy'+2y=e^x$$
$\textsf{divide thru by x}$
$$y' +\frac{2}{x}y=\frac{e^x}{x}$$
$\textsf{Find u(x)}$
$$\displaystyle u(x)=\exp\int\frac{2}{x} \, dx=e^{\ln x^2}=x^2$$
$\textsf{so far anyway..}$

 

[karush]

$\textsf{find the general solution}$
$$xy'+2y=e^x$$
$\textsf{divide thru by x}$
$$y' +\frac{2}{x}y=\frac{e^x}{x}$$
$\textsf{Find u(x)}$
$$\displaystyle u(x)=\exp\int\frac{2}{x} \, dx=e^{\ln x^2}=x^2$$

$\textsf{multiply thru with $x^2$}$
$x^2 y' +2xy=xe^x$
$(x^2 y)'=xe^x$
$\textsf{integrate}$
$\displaystyle x^2 y=\int xe^x \, dx =e^x(x-1)+c$
$\textsf{divide thru by $x^2$}$
$\displaystyle y=\frac{e^x}{x}-\frac{e^x}{x^2}+\frac{c}{x^2}$
hopefully